When are tracial states continuous in the strict topology?

Question

Let $A$ be a separable (non-unital) C*-algebra. Let $\tau: A\to\mathbb{C}$ a tracial state. Consider the strict topology on $A$, i.e. $a_n\to a$ iff $a_nx\to ax$ and $xa_n\to xa$ for all $x\in A$. It is clear that this is the same as the usual norm-topology when $A$ has a unit.

My question is:

When is $\tau$, restricted on the unit ball of $A$, continuous in the strict topology?

In other words: Under what conditions is it true that every norm-bounded sequence $a_n\in A$, with $a_nx~,~ xa_n\to 0$ for all $x\in A$, satisfies $\tau(a_n^*a_n)\to 0$?

Is this always possible? If not, I would be very glad to see a counterexample. If it turns out to be wrong in general, I would also like to know about the case that $A$ has a unique tracial state and is algebraically simple. Looking at that case would be sufficient for my purposes.

Answer 1

Go to the GNS representation. There is a cyclic vector $v$ such that $\tau(x) = \langle \pi(x)v,v\rangle$ for all $x \in A$. Now if $\|a_nx\| \to 0$ in $A$ for every $x \in A$ then $\|\pi(a_nx)v\|\to 0$ for all $x \in A$. Since $v$ is cyclic and $(a_n)$ is bounded, this implies that $\pi(a_n)v \to 0$, so $\tau(a_n^*a_n) = \|\pi(a_n)v\|^2 \to 0$.